Feb 3, 2015 · We can use the Sum-Product method. Bring everything to one side: x^2-x=12->x^2-x-12=0 We now have a quadratic equation of the form ax^2+bx+c=0 where a=1, b=-1 and c= …

Jun 1, 2013 · 3. You might not understand the parts of a fraction nor the notation: 2/x=12. The 2 is the numerator, and tells the number of parts of the whole that you have. The line is the, …

Feb 6, 2016 · x^2-x-12 = (x-4)(x+3) Look for a pair of factors of 12 that differ by 1. The pair 4, 3 works. Hence we find: x^2-x-12 = (x-4)(x+3) In general: (x-a)(x+b) = x^2-(a-b)x-ab In our case …

Jun 19, 2015 · It is not necessary, but later steps are easier if we first make sure that we have 0 on one side. (Any constant will be ok, but 0 is easiest.) Partition the number line by finding …

Mar 9, 2016 · #x^2+x-12=0# This is in the form of a Quadratic equation (in form #ax^2+bx+c=0#) Where. #a=1,b=1,c=-12 ...

Oct 25, 2015 · x=1 First, rearrange the equation like this: 2^(2x) + 2^(x + 2) - 12 = 0 (2^x)^2 + 4(2^x) - 12 = 0 Now, treat this like a quadratic equation by substituting 2^x=s: s^2+4s-12=0 …

\[{x}^{2}-x=12\] +. > < ...

Nov 26, 2015 · The you need to get one #x# on its own on one side of = and everything else on the other side. This is how you do it for your question: When multiplying, change what you …

Jan 15, 2016 · #x^2# can only be #x xx x# so we have a starting point of: #color(blue)((x+?)(x+?))# '~~~~~ #color(blue)("Consider the constant of "-12)# This is a …

Feb 8, 2017 · x^2-x-12>0" "=>" "x<"-"3 uu x>4. Since the right-hand side (RHS) is already 0, we start by factoring the left-hand side (LHS): " "x^2-x-12>0 =>(x-4)(x+3)>0 In its factored form, …